Chemical Equilibrium

By Doba Jackson, Ph.D.

Outline of Chpt 5• Gibbs Energy and Helmholtz Energy• Gibbs energy of a reaction mixture

(Chemical Potential)• Gibbs energy of mixing• Calculating Gibbs energy of reactions• Equilibrium constant• Equilibrium reactions with phase mixtures

Introduction to Free Energy

Gibbs Free Energy Change (∆G)

Entropychange

∆G =

Enthalpy ofreaction

Temperature(Kelvin)

∆H ∆S- T

∆G < 0 Process is spontaneous

∆G = 0 Process is at equilibrium

∆G > 0 Process is non-spontaneous

Enthalpy term islarger than Entropy

Entropy term islarger than Enthalpy

Types of Expansion Work• Irreversible Process

exw = -P ΔV

• Reversible Processf

i

V

V

nRTw = - dV

V

Comparison of work done upon expansion of an Ideal gas

0revdw dw

0revdq dq

ReversibleIrreversibledU dq dw

revdw dw

An irreversible process produces less work and heat than a reversible process

U is a state function. It does not depend on the path(reversible or irreversible)

Two Isothermal processes

0revdq dq

Isothermal, Reversible

Isothermal, Irreversible

revdw dw

An irreversible process produces less work and less heat than areversible process

revrev

revrev

dwdq

dwdq

dwdqdU

dwdqdU

0

0

0 dqdqrev

dqdqrev

The inequality of Clausius

0revdq dq

Isothermal, Reversible

Isothermal, Irreversible

revdq dq

T T

dqdqrev 0 dqdqrev

revrev dqTdST

dqdS ;

dqTdS Clausius Inequality

Reversible, NonspontaneousReversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initialstate leaving all thermodynamic variables for the universe (system + surroundings) unchanged.

A truly reversible change will:

- Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change

IrreversibleReversible

Irreversible, SpontaneousIrreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initialstate will change some thermodynamic variables of the universe.

A truly irreversible change will:

- Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change

IrreversibleReversible

Clausius Inequality leads to equations for spontaneity

0revdw dw

0revdq dq

ReversibleIrreversible

revdq dq

T T

dqdS

T

Constant Volume

TdS dq

VTdS dq TdS dU

Helmholtz Energy

dA = 0; Equilibrium

dA < 0; Irreversible, Spontaneous

TdSdUdA

STUA sSpontaneou-Non;0A

Clausius Inequality leads to equations for spontaneity

Reversible

IrreversibleConstant Pressure

TdS dq

pTdS dq TdS dH

Gibbs Energy

dG = 0; Equilibrium

dG < 0; Irreversible, Spontaneous

STHG

TdSdHdG

sSpontaneou-Non;0G

Properties of the Gibbs Energy

G H TS ( )dG dh d TS

( ) ( )dG d U PV d TS

( ) ( )dG dU d PV d TS

dG dU PdV VdP TdS SdT

dU q w dU TdS PdV

q TdSw PdV

dG TdS PdV PdV VdP TdS SdT

dG VdP SdT

Gibbs energy is a function of only pressure and temperature

Properties of the Gibbs Energy

dG VdP SdT

Gibbs energy is a function of only pressure and temperature

P

GS

T

T

GV

P

T P

G GdG = dP - dT

P T

Meaning of Gibbs energy

• The Gibbs energy is the maximum work (non-expansion) that can be performed at a given state (T,P).

• Because S is always >0, G will decrease as temperature is increased.

• As S increases, the slope of G verses T becomes more negative (see graph).

P

GS

T

P

GS

T

Meaning of Gibbs energy• The Gibbs energy is the

maximum work (non-expansion) that can be performed at a given state (T,P).

• Because V is always >0, G will increase as pressure is increased.

• As V increases, the slope of G verses P becomes more positive (see graph).

T

GV

P

Calculate the change in Gibbs Energy when pressure increases

dG VdP SdT

If temperature is constant:

dG VdP2

1

P

PG VdP

If volume change is small (liquid, solids):

G V P If volume change is not small (Gases):

2

1

PG = nRTLn

P

Problem:Calculate the entropy and enthalpy changes for this chemical reaction at 298K .

CO2 (g) + H2O (l) ------------ C6H12O6 (s) + O2 (g) 6 6 6

ΔfH= (CO2) = -393.5 kJ/mol

ΔfH= (H2O) = -285.8 kJ/mol

ΔfH= (C6H12O6) = -1273.1 kJ/mol

ΔSm= (CO2) = 213.8 J/mol

ΔSm= (H2O) = 70.0 J/mol

ΔSm= (C6H12O6) = 209.2 J/mol

ΔSm= (O2) = 205.2 J/mol

Sec 6.4-6.6: Gibbs energy of a mixing (Chemical Potential)

Chemical Potential

• Chemical Potential (μ)- is a measure of the potential that a substance has to undergo a change in state or composition

• Molar Gibbs energy (Gm)- is the maximum energy that a system has available for non-expansion work.

For all Pure substances: μ = Gm

Partial Molar Quantities

• Partial Molar Volume- the change in volume of a substance per mole of volume added to the larger mixture.

*, ,

jj P T n

VV

n

A A B BdV = V dn + V dn , , B

AA P T n

VV

n

, , A

BB P T n

VV

n

• Partial Molar Gibbs- the change in Gibbs energy of a substance per mole of volume added to the larger mixture.

, , '

jj P T n

G

n

A A B BdG = dn + dn , , B

AA P T n

G

n

, , A

BB P T n

G

n

Partial Molar Gibbs Energy

A A B BG = n + n This assumes the temperature and pressure is constant.

, , B

AA P T n

G

n

, , A

BB P T n

G

n

A A B BdG = VdP - SdT + n + n Fundamental Equation which includes composition

,max A A B Bw = n + nadd The Gibbs energy is the maximum additional non-expansion work. This depends on composition.

Thermodynamics of Mixing• Why does mixing occur spontaneously?

2

1

PG = RTLn

Pm From Second Law

2,2 ,1

1

PG = RTLn

Pm mG

22 1

1

P= RTLn

P Gm = μ

P= RTLn

P

PΘ = 1 bar

= RTLn P P = in bar units

This is the chemical potentials of the pure gasses before mixing

Why does mixing occur spontaneously

= RTLnA A AP

= RTLnB B BP

A A B BG = n + n Mixing A & Bat a constant T, P

A BG = RTLn n + RTLn ninit A BP P

A BG = RTLn n + RTLn nfinal A A B BP P

mix final initialG G G

A B A BG = RTLn n + RTLn n RTLn n + RTLn nmix A A B B A BP P P P

-mix A A B B A BG RTLnP n RTLnP n RTLnP n RTLnP n

-mix A A B B A BG RTLnP n RTLnP n RTLnP n RTLnP n

A Bmix A B

P PG RTLn n RTLn n

P P

Dalton’s Law of Partial Pressures

A AP P AA

P

P

mix A A B BG RTLn n RTLn n

AA

T

n

n

A A Tn n

mix A A T B B TG RTLn n RTLn n

mix T A A B BG n RT Ln Ln

Remember this equation

Enthalpy and Entropy of mixing Gases

mix T A A B BG n RT Ln Ln

For a Perfect Gas:

0mixH

S mix T A A B Bn R Ln Ln

Problem 5.8A container of volume 5.0 dm3 that is divided into two equal compartments. One side contains H2 gas and the other contains N2 gas in equal molar anounts. Both containers are at 25ºC and 1 atm. Calculate the Gibbs energy and Entropy of mixing when the partition is removed.

Sec 6.4-6.6: Gibbs energy of a chemical reaction

Standard rxn Gibbs energies (ΔrGº) and

Standard Gibbs energies of formation (∆fGº)

H2O(l)H2(g) + ½ O2(g)

ReactantsProducts

∆rG° = { (1)∆fGo(H2O)} – {(1)∆fG

o(H2) + (.5)∆fGo(O2)}

Reaction Gibbs energies (ΔrGº) can be determined by the difference of the product Gibbs energies of formation and the reactant Gibbs energies of formation.

0 0 0r

Products Reactants

Δ = v - vf fG G G

Example:

Sec 6.8-6.10: Gibbs energy of a chemical reaction mixture